Setting up a triple integral using cylindrical & spherical coordinates

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Homework Help Overview

The discussion revolves around setting up a triple integral for a region defined inside a sphere and between two planes, specifically using cylindrical and spherical coordinates.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the limits of integration for both cylindrical and spherical coordinates, questioning the correctness of upper and lower limits for the variables involved.
  • Some participants express uncertainty about the presence of a coefficient in front of the integral and seek clarification on the dependency of variables on angles.

Discussion Status

Participants are actively engaging with the problem, providing feedback on each other's attempts and clarifying concepts related to the limits of integration. There is a focus on understanding the relationship between the geometric setup and the mathematical representation in different coordinate systems.

Contextual Notes

Some participants note the need to express the plane equations in spherical coordinates and discuss how these expressions affect the limits of integration. There is an acknowledgment of the dependency of certain variables on angles, which adds complexity to the setup.

calcuseless
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Homework Statement



Inside the sphere x2 + y2 + z2 = R2 and between the planes z = [tex]\frac{R}{2}[/tex] and z = R. Show in cylindrical and spherical coordinates.

Homework Equations



[tex]\iiint\limits_Gr\,dz\,dr\,d\theta[/tex]

[tex]\iiint\limits_G\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta[/tex]

The Attempt at a Solution



[tex]2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_?^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta[/tex]
Is my upper limit for r correct? How do I find the lower limit for z?

[tex]\int_0^{2\pi}\int_0^{\frac{\pi}{3}}\int_?^R\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta[/tex]
How do I find the lower limit for rho?
 
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calcuseless said:

Homework Statement



Inside the sphere x2 + y2 + z2 = R2 and between the planes z = [tex]\frac{R}{2}[/tex] and z = R. Show in cylindrical and spherical coordinates.

Homework Equations



[tex]\iiint\limits_Gr\,dz\,dr\,d\theta[/tex]

[tex]\iiint\limits_G\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta[/tex]

The Attempt at a Solution



[tex]2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_?^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta[/tex]
Is my upper limit for r correct? How do I find the lower limit for z?

Yes the upper limit is OK. Isn't the equation of the plane z = R/2? That would be z on the lower surface.

[tex]\int_0^{2\pi}\int_0^{\frac{\pi}{3}}\int_?^R\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta[/tex]
How do I find the lower limit for rho?

Again, the equation of that plane is z = R/2. Express that in spherical coordinates and solve it for [itex]\rho[/itex]. You will find that [itex]\rho[/itex] depends on [itex]\phi[/itex], but that's OK.
 
LCKurtz said:
You will find that [itex]\rho[/itex] depends on [itex]\phi[/itex], but that's OK.

Ah, I didn't understand that concept. Thank you!

[tex]2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_\frac{R}{2}^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta[/tex]
Is there supposed to be a 2 out in front? I don't know why I put it there and I don't think it is needed for this problem.
 
calcuseless said:
[tex]2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_\frac{R}{2}^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta[/tex]
Is there supposed to be a 2 out in front? I don't know why I put it there and I don't think it is needed for this problem.

I don't know why you put it there either. Go with your instincts. If you don't think it should be there, don't put it there. :cool:
 

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